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A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. at time t1 = 5 s, it is at point (4.00 m, 3.00 m) with velocity (1.50 m/s)j and acceleration in the positive x direction. at time t2 = 10 s, it has velocity (-1.5 m/s)i hat and acceleration in the positive y direction. what are the coordinates of the center of the circular path?

Respuesta :

Initial velocity at t = 5 s

[tex]v = 1.5 \hat j[/tex]

final velocity at t = 10 s

[tex]v = -1.5 \hat i[/tex]

so the time taken by the object to revolve by 270 degree is t = 10 - 5 = 5s

so here we can find the angular speed

[tex]\omega = \frac{\theta}{t}[/tex]

[tex]\omega = \frac{\frac{3\pi}{2}}{5}[/tex]

[tex]\omega = 0.94 rad/s[/tex]

now we know that

[tex]v = R\omega[/tex]

[tex]1.5 = 0.94 * R[/tex]

[tex]R = 1.59 [/tex]

so the coordinates of the center will be

[tex]x = 4 + 1.6 = 5.6 m[/tex]

[tex]y = 3 m[/tex]

so the coordinates of center is (5.6 m,3 m)

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